Relation between covariant derivative and differential Let $\frac{\hat{D}}{dt}$ be the covariant derivative via the Levi-Civita connection. For the definition of this, my reference is Lee's "Riemannian manifolds", Lemma 4.9. For the present context, this says that: for some vector field $X_t$ along a curve $u(t)$, the covariant derivative of $X_t$ is $\frac{\hat{D}}{dt} X_t = \nabla_{\dot{u}(t)}X(t)$, where $\nabla$ in our case is the Levi-Civita connection.
Furthermore, let $V$ be a vector space, considered as a manifold.
Short version of the question
Given some smooth curve $p(t)$ in V, can we make sense of $\frac{\hat{D}}{dt} p(t)$, and does it hold that $\frac{\hat{D}}{dt} p(t) = \frac{d}{dt} p(t)$?
I know that the tangent spaces of a vector space can be identified with the space itself.
Longer version
I'm trying to understand the following argument (from this paper).
Let $e_{\epsilon}(t)$ be a curve in a Lie group, considered either as a function of $\epsilon >0$ or $t > 0$. This curve fulfills $e_{0}(t) = e$ (the exact definition of the curve, from a differential equation, can be found on page 5 in the paper).
Let $\frac{\hat{D}}{dt}$ be the covariant derivative via the Levi-Civita connection.
In the paper, it is then established that
$$\frac{\hat{D}}{dt} \frac{d}{d\epsilon} e_{\epsilon}(t) = T_e L_{e_{\epsilon}(t)}\dot{v}(t) + \epsilon \frac{\hat{D}}{d\epsilon}(T_e L_{e_{\epsilon}(t)}\dot{v}(t)),$$
which implies that
$$ \left.\frac{\hat{D}}{dt} \left(\frac{d}{d\epsilon} e_{\epsilon}(t) \right|_{\epsilon=0}\right) = \left(\left.\frac{\hat{D}}{dt} \frac{d}{d\epsilon} e_{\epsilon}(t)\right) \right|_{\epsilon=0} =\dot{v}(t).$$
The thing I don't understand is the following conclusion from the above;
$$\left. \frac{d}{d t} \left(\frac{d}{d\epsilon} e_{\epsilon}(t) \right|_{\epsilon=0}\right) = \dot{v}(t).$$
My thoughts:
since $\left.\frac{d}{d\epsilon} e_{\epsilon}(t) \right|_{\epsilon=0}$ is a curve in the Lie algebra, they seem to use that the covariant derivative reduces to the differential  if the input is a curve on a vector space (in this case the Lie algebra) - but I haven't been able to establish that fact.
Thanks
 A: Ok, so I arrived at the following conclusion: for the context in my question, the covariant derivative wrt. the Levi-Civita connection reduces to the differential. 
My proof:
By lemma 5.2 in Lee's "Riemmanian manifolds", the covariant derivative wrt. the Levi-civita connection fullfills the following. Let $U(t),W(t)$ be vector fields along some curve on V, then
$$\frac{d}{dt}\langle U(t),W(t)\rangle = \langle \frac{\hat{D}}{dt}U(t),W(t)\rangle + \langle U(t),\frac{\hat{D}}{dt}W(t)\rangle.$$
Furthermore, for any vector space we have the following identity (see theorem 8.4 in Sternberg's advanced calculus part 3),
$$\frac{d}{dt}\langle U(t),W(t)\rangle = \langle \frac{d}{dt}U(t),W(t)\rangle +  \langle U(t),\frac{d}{dt}W(t)\rangle,$$
where we recall that each tangent space of a vector space can be identified with the vector space itself.
Choosing constant $W(t) = w$ for arbitrary $w \in V$, we get
$$\langle \frac{d}{dt}U(t),w\rangle = \langle \frac{\hat{D}}{dt}U(t),w\rangle ,$$
so $$\frac{d}{dt}U(t) = \frac{\hat{D}}{dt}U(t).$$
... and adjusting this answer to the context of the "long version" above:
Let $\epsilon \mapsto e_{\epsilon}(t)$ be the curve in the lie group G. Then $t \mapsto \left.\frac{d}{d\epsilon}\right|_{\epsilon=0}e_{\epsilon}(t)$ is a curve in the Lie algebra. $\frac{d}{dt} (\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}e_{\epsilon}(t))$ can then be considered a vector field on G, along the constant curve $\gamma(t) = e$.
Using the arguments above, it holds that $\frac{\hat{D}}{dt} (\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}e_{\epsilon}(t)) = \frac{d}{dt} (\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}e_{\epsilon}(t))$.
... corrections and comments will be greatly appreciated!
